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will be equal to the rectangle under the segments of the hypotenuse; and that, conversely, if this be the case, the angle BAC will be a right angle. (See the proof of Cor. 2.)
Among the different proofs which have been invented of this celebrated theorem, there is one of no little elegance, which has the advantage of pointing out in what manner the squares of the two sides may be dissected, so as to form, by juxta-position of their parts, the square of the hypotenuse. It will be readily apprehended from the following outline. BAC is a triangle, having the angle at A a right o angle: upon the hypotenuse BC is described the square BCDE, and through the points D, E, straight lines are drawn parallel to B A, CA, to meet one another in K, and A C, A B produced, in the points F, G. Then it may be shewn that the four right-angled triangles, upon the sides of the square BD, are equal to one another; and, therefore together equal to twice the rectangle BA, AC also, that the figure AFKG is a square, and equal to the square of BA+AC: and hence it easily follows that the square of BC is equal to the squares of B A, A C (32. and ax. 3.). To shew the dissection of the squares, Ee and Cc are drawn paral lel to A B, to meet B b, which is drawn parallel to A C, in the points c, e: then Cb is equal to the square of A B, and Eb to the square of A C; and the former is divided into two, and the latter into three parts, which may be placed (as indicated by the divisions of B D,) so as to fill up the space BCDE, which is the square of BC.
PROP. 37. (EUC. ii. 12. and 13.) In every triangle, the square of the side which is opposite to any given angle, is greater or less than the squares of the sides containing that angle, by twice the rectangle, contained by either of these sides, and that part of it, which is intercepted between the perpendicular let fall upon it from the opposite angle, and the given angle: greater, when the given angle is greater than a right angle, and less, when it is less.
Let A B C be any triangle, and C one of its angles; and from the angle A to the opposite side BC, or BC produced, let there be drawn the perpendicular A D: the square of A B shall be greater or less than the squares of A C, C B, by twice the rectangle BC, CD; greater, if C be greater than a right angle, and less if it be less.
When C is greater than a right angle, the opposite side, A B, and the perpendicular, A D, must lie upon different sides of A C; since, otherwise, in the triangle ACD, one of the angles would be a right angle, and another greater than a right angle, which is impossible (8.). For the like reason, when C is less than a right angle, the opposite side, A B, and the perpendicular, AD, must lie upon the same side of A C.
Therefore, according as C is greater or less than a right angle, the line BD will be the sum, or the difference of BC, CD and (32. and 33.) the square of B D will be greater or less than the squares of B C, C D by twice the rectangle BC, CD. Add to each the square of AD: therefore, the squares of B D, A D will be greater or less than the squares of B C, CD, A D, by twice the rectangle B C, C D. But the square of A B is equal (36.) to the squares of BD, AD, and the square of AC is equal to the squares of CD, A D. Therefore, the square of AB will be greater or less than the squares of B C, AC by twice the rectangle B C, CD. Therefore, &c.
Cor. Any angle of a triangle is greater or less than a right angle, according as the square of the side opposite to it is greater or less than the squares of the sides by which it is contained.
or to the base produced; the difference of the squares of the sides shall be equal to the difference of the squares of the segments of the base, or of the base produced.
Let ABC be a triangle, and from the vertex A to the base BC, or B C produced, let there be drawn the perpendicular AD: the difference of the squares of AB, A C, shall be equal to the difference of the squares of BD, CD. For the square of BD is as much greater than the square CD, as the squares of B D, A D together are greater than the squares of C D, A D together, that is (36.), as the square of AB is greater than the square of A C.
We may remark that, if the triangle be isosceles, the segments of the base will be equal; and, that, in other cases, the greater segment of the base is always adjoining to the greater side. (12. Cor. 1, and 2.)
In an isosceles triangle, if a straight line be drawn from the vertex to any point in the base, or in the base produced, the square of this straight line shall be less or greater than the square of either of the two sides, by the rectangle under the segments of the base, or of the base produced.
Let ABC be an isosceles triangle, having the side A B equal to the side AC, and from the vertex A to any point D in the base, or in the base produced, let there be drawn the line AD: the square of AD shall be less or greater than the square of A C, by the rectangle B D, D C.
Bisect BC in E, (post. 3.) and join AE. Then, DC is equal to the sum of DE and EC; and because B E is equal to EC, BD is equal to the difference of DE and E C. Therefore the rectangle B D, DC is (34.) equal to the difference of the squares of DE, EC. But the line AE is perpendicular to BC, because the base BC of the isosceles triangle is bisected in E (6. Cor. 3.). Therefore, (38.) the difference of the squares of DE, EC, is equal to the difference of the squares of AD, AC. Therefore (ax. 1.) the rectangle BD, DC, is equal to the difference,
Let ABC be any triangle, and from the vertex A to D, the middle point of the base, let there be drawn the line AD. The squares of A B, A C, shall be together double of the squares of B D, DA.
B DEC BDCE
From the point A to B C, or B C produced, draw the perpendicular A E (12.). Then BE is equal to the sum of BD, DE; and, because B D is equal to DC, EC is equal to the difference of B D, DE. Therefore, (35.) the squares of B E, EC are together double of the squares of BD, DE. And the square of EA, taken twice, is double of the square of E A. Therefore the squares of BE, EA, CE, EA, are together double of the squares of BD, DE, EA; that is (36.), the squares of BA, AC _are_together double of the squares of BD, D A.
When ACB is a right angle, the perpendicular AE coincides with the side A C. Therefore B D is equal to DE, and the square of B E is double of the squares of BD, DE (29. Cor. 2.) ; also the square of BE (36. Cor. 1.) is equal to the difference of the squares of B A, A E, that is, of BA, AC; and hence the squares of B A, A E, that is, of B A, A C, are together double of the squares of B D, D A, as before. Therefore, &c.
together greater than the squares of Operations so simple require no further AC, BD, by four times the square of EF.
Join BF, FD. Then, because the base AC of the triangle BAC is bisected in F, the squares of A B, BC are (40.) equal to twice the squares of B F, A F: and for the like reason the squares of CD, DA are equal to twice the squares of D F, AF: therefore, the squares of A B, B C, C D, DA are together equal to twice the squares of BF, DF, together with four times the square of A F. But, because the base BD of the triangle FBD is bisected in E, the squares of BF, DF, taken twice, are (40.) equal to four times the squares of BE, EF: therefore the squares of A B, BC, CD, DA are together equal to four times the squares of A F, BE. together with four times the square of EF; that is, to the squares of AC, BD, (39. Cor. 2.) together with four times the square of E F.
Cor. In a parallelogram, and in a parallelogram only, the points E and F coincide, because (22.) the diagonals bisect one another: therefore, in a parallelogram, and in a parallelogram only, the squares of the diagonals are together equal to the squares of the four
Upon a piece of paper for a plane, with a pen,* a ruler, and a pair of compasses, it is evident, that, first, a straight line may be drawn from any one point to any other point; 2ndly, a terminated straight line may be produced to any length in a straight line; 3dly, from the greater of two straight lines, a part may be cut off equal to the less; and 4thly, a circle may be described from any centre, and with any distance from that centre.
Although the paper be not an exact plane, nor the pen such as may serve to draw an exact line, these defects admit of being removed to any required degree, and do not, in the least, affect the accuracy of our conclusions, with regard to exact lines and planes. An edge which is a right line, or nearly so, (because it is the common section of two planes, see Book IV.) may be obtained by doubling over a piece of paper upon itself, and a right angle (def. 10.) by doubling over this edge upon itself. These are both useful upon occasion, especially the right angle, which is so frequently required in geometrical constructions that a case of instruments is commonly provided with one. Among practical mechanics it is known under the name of the square. Parallel lines also occur so frequently, that it is convenient to have a ruler expressly for drawing them, called a parallel ruler. This may be made of two or three different forms, the best of which is that of a flat ruler running
upon two equal rollers.
PROP. 42. Prob. 1. (Euc. i. 1.)
as centres, with the com-
Then, because CA, C B are each of them equal to A B (def. 24.), they are (ax. 1.) equal to one another, and the triangle CAB is equilateral. Therefore, an equilateral triangle has been described upon the given finite straight line AB, which was required to be done.
PROP. 43. Prob. 2. (Euc. i. 10.) To bisect a given finite straight line A B.
Upon either side of A B (42.) describe an equilateral triangle: join the vertices or summits C,D, and let ČD cut AB in E.
CBD have the three sides of the one Then, because the triangles CAD, equal to the three sides of the other, each to each, (7.) they are equal in every respect, and the angle ACD is equal to BCD. Therefore, the triangles ACE, BCE, having two sides of the one equal to two sides of the other, each to each, and the included angles equal to one another, are equal in every respect (4), and A E is equal to E B. Therefore, &c.
Cor. By the same construction, a straight line may be drawn, which shall bisect any given straight line at right angles.
N. B. It is evident, from 6. Cor. 4., that the same end will be obtained by joining the point of intersection C of any two equal circles described from the centres A, B, and cutting one another section D of any other two equal circles above A B, with the point of interdescribed from the same centres, and cutting one another either above or below A B: for C, D will be the vertices of isosceles triangles upon the same base. This observation may be of use when one of the triangles is already described, as in the first method of Prob. 4.
PROP. 44. Prob. 3. (Euc. i. 11.) To draw a straight line at right angles to a given straight line AB, from a given point C in the same.j
In CA take any from the centre C',
First method. point A, and make CB equal to CA: upon AB describe (42.) the equilateral triangle DA B, and join D C.
Then, because the triangles DCA, DCB have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle DCA is equal to DCB; and they are adjacent angles; therefore each of them is a right angle (def. 10.), and D C is at right angles to A B.
Second method. Take any point D which is not in AE, and from the centre D, with the radius DC describe the circle AC E, cutting the line AB a second time in the point A: join AD, and produce it to meet the circumference in E, and join E C. Then because in the triangle E CA, the line D C drawn from the angle C to the middle of the opposite side equal to half that side, the angle C is (19. Cor. 4.) a right angle, and CE is at right angles to A B. If the points A, C, coincide, DC is (12. Cor. 3.) at right angles to
with the radius. C D, describe a circle cutting A B in the points A, B: bisect (43.) AB in E, and join CE.
Then, because the triangles CEA, CEB have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle CEA is equal to CEB, and (def. 10.) C E is at right angles to A B.
Second method. Draw from C to (43.) in D: from the centre D, with the AB any straight line C A : bisect CA radius DA or DC, describe a circle, cutting AB in a second point E, and join CE. Then, for the same reason,
as in the second method
points A, B, and from A, B, as centres, Third method. In A B take any two with the radii A C, BC, describe circles second time in F: join cutting one another a CF, and let CF cut AB in E. Then, because ACF, BCF, are isosceles triangles upon the same base CF, the line AB which joins their summits, bisects the base at right angles (6. Cor. 4.); that is, CE is at right angles to AB.
When CA is equal to CB, (which may happen, when the points A, B are on different sides of E,) the last of these three methods is, in practice, nearly the
same as the first. The last two methods point C falls near the edge of the paper. are applicable to the case, in which the
PROP. 46. Prob. 5. (Euc. i. 9.) To bisect a given rectilineal angle BAC. In A B take any point B: make A C equal to AB: join BC: upon B C describe (42.) the equilateral triangle BD C, and join A D. Then, because the triangles A B D, ACD have the three sides of the one equal to the three sides of the other, each to each, (7.) the angle B A D is equal to the angle CAD.
Cor. By repeating this process with the halves, quarters, &c. of the given angle, it may be divided into four, eight, &c., equal parts.
There is no construction in Plane Geometry, i. e. no construction practicable with the assistance of the right line and the circle only, by which a given angle may be divided into three, five, &c. equal parts. It is true that, by the description of other lines which are plane curves, (as the hyperbola, conchoïd, and cissoïd,) any given angle may be trisected, or divided into three equal parts; but such constructions, not being effected by the two species of lines which are the subject of these Elements, are for certain reasons (to be found in the application of Algebra to Geometry) said to be out of the range of Plane Geometry. There are, how. ever, some particular cases in which, by means of the circle, an angle may be divided into three, five, and other numbers of equal parts. Such is the division of a right angle into three equal parts. In this case we are already aware, that, since an equilateral triangle has its three angles equal to one another, and the three taken together equal to two right angles, each of them is two thirds of a right angle. The problem, therefore, will be solved by describing an equilateral triangle upon either of the legs, taken of any length at pleasure. In like manner, by means of an isosceles triangle, each of whose base-angles is double of the vertical angle, (see book iii.) and is, therefore, four-fifths of a right angle, Plane Geometry enables us to divide a right angle into five equal parts. But these are only particular cases, and indicate no general process.
PROP. 47. Prob. 6. (Euc. i. 23.) At a given point A, in a given straight line AB, to make an angle equal to the given rectilineal angle Č.
From the centre C, with any radius, describe a circle, cutting the sides of the given angle in the points D E, and
N. B.—In the figure, equal circles are, centres A, C, the former cutting BC a in the first place, described from the second time in B; and the point D is then determined by describing a circle from the centre A with the radius B C.
This construction is equally short, in practice, with that of Prop. 47.; and when it is required to obtain the point D at as great a distance as possible from A, may be preferred, the line A C being drawn more obliquely for that purpose.
PROP. 49. Prob. 8. (Euc. vi. 9). To divide a given straight line, A B, into any number of equal parts.
Let it be required to divide A B into five equal parts. Draw A C, making any angle with A B, and taking any distance M, set it off upon A C five times from A to C: join C B, and through the points of division in A C, viz, D, E, &c. draw (48.) parallels to C B, cutting A B in the points F, G, &c. AB shall be divided at the latter points into five equal parts, Through F and G draw (48.) F H and GK, each of them parallel to A C. hen, because FC is a parallelogram, FH is equal to DC (22.); and for the like reason G K is equal to ED: but ED, D C, are equal to one another; therefore (ax. 1.), FH is equal to G K. But FH, GK, are sides of the